During the nuclear explosion, one of the products is ⁹⁰Sr with half life of 6.93 years. If 1 μ g of ⁹⁰Sr was absorbed in the bones of newly born baby in placed of Ca, how much time, in years, is required to reduce it by 90% if it is not lost metabolically? (Round off to the nearest integer)

Solution

Chemistry ma’am · 4 steps

Reduced by 90% means only 10% is left, so set N₀/N = 10 and the half-life does the rest.

1. Decay constant. Radioactive decay is first order, so λ = 0.693/t_(1/2) = 0.693/(6.93 y) = 0.1 y⁻¹.
2. What is left. A 90% reduction leaves 10% of the strontium. If N₀ = 1 μg, then N = 0.1 μg and N₀/N = 10. The starting mass does not matter, only the ratio.

   t = 2.303/λ log N₀/N

3. Put in numbers. t = 2.303/0.1 log 10 = 23.03 × 1 = 23.03 y.
4. So. The time needed is about 23.03 years, which rounds to 23.

The common slip: 1. Treating 90% as the amount left uses log 1/0.9 and gives about 1 year, not 23. After a 90% reduction, only 10% remains.

The idea to keep. For first-order decay, use the fraction that remains, not the fraction that decayed. Falling to 1/10 always takes 2.303/λ, about 3.32 half-lives.

Answer from the NTA final answer key. Solution by PrepWiser.